Line Integral and Green's Theorem

[Mrs.Fling]

Hi! New here...I've been struggling with this problem for a couple of days. I've tried applying several examples, but I don't see where I'm going wrong. Here is the exact problem text:
By making an appropriate choice for the functions P(x,y) and Q(x,y) that appear in Green's Theorem in a plane, show that the integral of x-y over the upper half of the unit circle centered on the origin has the value -2/3. Show the same result by direct integration in Cartesian coordinates.

Ok, so let's call the Green's Theorem part, part (a) and the direct integration part, part (b).

For part (a), I really don't know how to pick P and Q. It seems I have this function, f(x,y) = x-y. And Green's theorem starts with a vector. Is my vector F=xi - yj? I also thought about taking the gradient of f(x,y), which is just =i - j and I'm not really sure what that does for me.

For part (b), I've applied many examples. Most use parametrization and set x=cos t and y = sin t. The magnitude of this is just 1, so then my integral becomes cos t -sin t dt? But then I just get 0. I have uploaded a pic of my work for this.

Any insight is appreciated! Thanks!

 

[galactus]

By using \(\displaystyle xdy-ydx\) and letting \(\displaystyle x=rcos(t), \;\ y=rsin(t), \;\ dx=rsin(t), \;\ dy=rcos(t)\), we have:

\(\displaystyle \displaystyle\int_{0}^{\pi}\int_{0}^{1}r^{2}(cos(t)-sin(t))drdt\)

 

[Mrs.Fling]

Understanding the problem

Hi Galactus,

A couple of questions....first, how are you able to paste equations from equation editor? When I use the "paste from Word" button, it looks like this:

or

Second question, if x = r cos(t), wouldn't dx = -rsin(t) dt + cos(t) dr? It seems to me like you just took the derivative of cos(t) to be sin(t)? Which the derivative of cos(t) is -sin(t). Maybe I'm missing something?

3rd question, the other students in my class are not taking the parametric approach. Instead, they are just doing I= x-y dydx. How do you know when to substitute in x = r cos(t) and when not to?

Lastly, would mind writing out your individual steps? I just didn't see how you got the rdrdt. My calculus is a bit rusty.

Thank you so much!

 

[galactus]

I did not use the equation editor for that. It is LaTex. I typed the code. Click on 'quote' in the upper right corner of this post to see the code I used to make it display in the nice math type.


Sorry for the misleading post. What I actually done was used your x-y. Then, let

\(\displaystyle x=rcos(t), \;\ y=rsin(t)\)

When integrating in polar, remember the "extra r".

So, we end up with the aforementioned integral.

To use rectangular,

\(\displaystyle \displaystyle \int_{-1}^{1}\int_{0}^{\sqrt{1-x^{2}}}(x-y)dydx\)

Either way, one gets -2/3.

You can use polar when there are circles and polar regions involved. I thought of polar right away when the problem mentioned a unit circle. This isn't parametric. I just used t instead of theta.

 

[Mrs.Fling]

Ok, thanks! I used LaTex in college. Didn't know this thing could do that...pretty cool.

 

[galactus]

Yes, many of the better math sites support LaTex. They would not be much account without it.

If you continue on the site, remember the code must be encased in \(\displaystyle tags.

You can see them by clicking on 'quote' on any post that utilizes it. The tags may vary from site to site. Some use [latex], others use \(\displaystyle , and some just use $\)\)